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How do you simplify surds, carry out the four operations with them, and rationalise a denominator?

Simplify surds, carry out the four operations with surds, expand brackets containing surds, and rationalise the denominator of a fraction (Higher tier).

A focused answer to the WJEC GCSE Mathematics Higher number content on surds, covering simplifying, the four operations, expanding brackets with surds, and rationalising the denominator.

Generated by Claude Opus 4.811 min answer

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  1. What this dot point is asking
  2. What a surd is
  3. Simplifying surds
  4. Adding, subtracting and multiplying
  5. Rationalising the denominator
  6. Why surds matter

What this dot point is asking

WJEC places surds in the Higher-tier number content. A surd is a root that cannot be written exactly as a fraction, so working with surds keeps answers exact rather than rounding. You must simplify surds, add, subtract, multiply and divide them, expand brackets that contain them, and rationalise denominators, including the conjugate case. Surds appear on the non-calculator Unit 1 and feed straight into the quadratic formula, Pythagoras and trigonometry with exact values, so they are a high-value Higher topic.

What a surd is

An irrational number cannot be written as an exact fraction; its decimal neither terminates nor recurs. The square root of any whole number that is not a perfect square is irrational, so 2,3,5\sqrt{2}, \sqrt{3}, \sqrt{5} are surds, but 9=3\sqrt{9} = 3 is not. Leaving an answer as a surd is leaving it exact, which is why WJEC questions say "give your answer in surd form" or "in the form aba\sqrt{b}".

Simplifying surds

The key move is to split out a square factor.

So 72=36×2=62\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}. Choosing the largest square factor finishes in one step; using a smaller one (72=4×18=218\sqrt{72} = \sqrt{4 \times 18} = 2\sqrt{18}) means you must simplify again.

Adding, subtracting and multiplying

Surds behave like algebraic terms: only like surds combine.

For addition and subtraction, simplify first so that matching surds appear, then add the coefficients: 12+27=23+33=53\sqrt{12} + \sqrt{27} = 2\sqrt{3} + 3\sqrt{3} = 5\sqrt{3}. For multiplication, multiply coefficients together and surds together: 25×32=6102\sqrt{5} \times 3\sqrt{2} = 6\sqrt{10}. When you multiply a surd by itself the root disappears: 7×7=7\sqrt{7} \times \sqrt{7} = 7. Expanding brackets follows the usual rules, for example (2+3)(1+3)=2+23+3+3=5+33(2 + \sqrt{3})(1 + \sqrt{3}) = 2 + 2\sqrt{3} + \sqrt{3} + 3 = 5 + 3\sqrt{3}.

Rationalising the denominator

Convention says a final answer should not have a surd in the denominator.

Why surds matter

Surds are how WJEC asks for exact answers in Pythagoras, trigonometry with the special angles (sin60=32\sin 60^\circ = \tfrac{\sqrt{3}}{2}) and the quadratic formula when the discriminant is not a perfect square. Keeping a value as 525\sqrt{2} rather than 7.077.07\ldots avoids rounding error that would otherwise compound through a multi-step problem, and WJEC's mark schemes specifically reward exact surd answers where they are requested on Unit 1.

Exam-style practice questions

Practice questions written in the style of WJEC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WJEC 20193 marksSimplify 48+27\sqrt{48} + \sqrt{27}, giving your answer in the form aba\sqrt{b}. (Higher, Unit 1, non-calculator.)
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Simplify each surd by taking out the largest square factor.

48=16×3=43\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3} and 27=9×3=33\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}.

Both now share 3\sqrt{3}, so add the coefficients: 43+33=734\sqrt{3} + 3\sqrt{3} = 7\sqrt{3}.

Markers award a mark for each correct simplification and a mark for the combined answer 737\sqrt{3}. Adding under the root to get 75\sqrt{75} is wrong; surds add only when the number under the root matches.

WJEC 20214 marksRationalise the denominator of 1053\dfrac{10}{5 - \sqrt{3}} and write your answer in the form a+b3c\dfrac{a + b\sqrt{3}}{c}. (Higher, Unit 1, non-calculator.)
Show worked answer →

Multiply top and bottom by the conjugate 5+35 + \sqrt{3}, because (53)(5+3)(5 - \sqrt{3})(5 + \sqrt{3}) is rational.

Denominator: (53)(5+3)=253=22(5 - \sqrt{3})(5 + \sqrt{3}) = 25 - 3 = 22.

Numerator: 10(5+3)=50+10310(5 + \sqrt{3}) = 50 + 10\sqrt{3}.

So 50+10322=25+5311\dfrac{50 + 10\sqrt{3}}{22} = \dfrac{25 + 5\sqrt{3}}{11} after dividing top and bottom by 22. Markers give marks for choosing the conjugate, the rational denominator 2222, the expanded numerator, and the simplified final form.

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